Question

Geometric Sequences and Series:

Answer 1

The video link doesn't work for me, but I assume this is the problem involving choosing between two payment types, one that immediately grants you some lump sum of cash, while the other starts off with $0.01 on the first day, $0.02 on the second day, $0.04 on the third, and so on, doubling per day.
Let $m_n$ be the amount of money earned on the $n$-th day. Then $m_n$ is a geometric sequence that satisfies

$\begin{cases}m_1=0.01\\m_n=2m_{n-1}&\text{for }n>1\end{cases}$

We can solve explicitly for $m_n$ in terms of the starting payment $m_1$:

$m_n=2m_{n-1}$
$m_{n-1}=2m_{n-2}\implies m_n=2^2m_{n-2}$
$m_{n-2}=2m_{n-3}\implies m_n=2^3m_{n-3}$
$\cdots$
$m_2=2m_1\implies m_n=2^{n-1}m_1$

So the amount of money earned on the $n$-th day is

$m_n=2^{n-1}\cdot0.01$

Denote the total amount of money earned over $n$ days by $M_n$. Then we can write

$M_n=\displaystyle\sum_{i=1}^nm_i=m_1+m_2+\cdots+m_{n-1}+m_n$

but since we have equivalent expressions for $m_i$ on any given day $i$, this is the same as

$M_n=\displaystyle\sum_{i=1}^nm_i=0.01+2\cdot0.01+\cdots+2^{n-2}\cdot0.01+2^{n-1}\cdot0.01$
$M_n=0.01(1+2+\cdots+2^{n-2}+2^{n-1})$

Let's call the sum on the right hand side $S_n$. Notice that
$S_n=1+2+\cdots+2^{n-2}+2^{n-1}$

$2S_n=2+2^2+\cdots+2^{n-1}+2^n$
$\implies S_n-2S_n=1-2^n$
$\implies-S_n=1-2^n$
$\implies S_n=2^n-1$

which means we end up with

$M_n=0.01(2^n-1)$


To answer part (D), you need to look no further than the formula for $M_n$. The question is basically asking what happens as $n$ gets arbitrarily large. It should be clear that $2^n$ grows without bound, so as $n\to\infty$, the amount of money you would get would diverge to infinity. So technically, you cannot calculate the amount because (1) there's only a finite amount of money to go around, and (2) infinity is not a "computable" number.
If, however, the scale factor used on your income was smaller than 1 - for example, say you were started with a million dollars on the first day, then your income got halved each day - then $m_n$ would eventually converge to 0, and on top of that, $M_n$ would converge to a finite number.